Question

$\sqrt{x^2+3x+4} + \sqrt{x+1} \ \textgreater \ 1,4$

Answer 1

     First we must pay attention to the existence conditions.

$\left \{ {{(I) x^2+3x+4 \geq 0 -\ \textgreater \ x \geq -1.5} \atop {(II)x+1 \geq 0-\ \textgreater \ x \geq -1}} \right. \\ \\ \boxed {E.C: x \geq -1}$
  
     Now we can proceed with calculations.

$\sqrt{(x+1.5)^2}+ \sqrt{x+1}\ \textgreater \ 1.4 \\ \sqrt{x+1}\ \textgreater \ 1.4-x-1.5 \\ ( \sqrt{x+1})^2\ \textgreater \ (-x-0.1)^2 \\ x+1\ \textgreater \ x^2+0.2x+0.01 \\ x^2-0.8x-0.99\ \textless \ 0$

$\Delta = (-0.8)^2-4.1.(-0.99) \\ \Delta = 0.64+3.96 \\ \Delta =4.6 \\ \\ x_{1}= \frac{0.8+ \sqrt{4.6} }{2} \\ \\ x_{2}= \frac{0.8- \sqrt{4.6} }{2}$
  
     Note that the second root no satisfies the existence condition, then it should not be in solution.

$\boxed { S=\frac{0.8+ \sqrt{4.6}}{2} }$

If you notice any mistake in English, please let me know, because I'm not native.